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Prove that if n is a perfect square, then n +2 is not a perfect square
 one year ago
 one year ago
Prove that if n is a perfect square, then n +2 is not a perfect square
 one year ago
 one year ago

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lgbasalloteBest ResponseYou've already chosen the best response.0
i suppose first step would be to assume n is perfect square
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
then i suppose i find n^2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
n + 2 would be m^2 + 2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
so now... i assume m^2 + 2 is a perfect square...
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
we have no 2 perfect square numbers with difference 2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
so i let m^2 + 2 = k^2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
@mukushla that's not really a proof.....
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
you may continue @igba
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
hmm i suppose the next step would be m^2  k^2 = 2 so (m+k)(mk) = 2 i don't think this proves anything...
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
actually that is :) \[m^2k^2=2\]\[(mk)(m+k)=2\]and this is impossible
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
well,,you've almost proved it.. now m and k both are integers/.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
and its k^2 m^2 = 2 and not otherwise..
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
now...integers huh...
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
\[m+k>mk\]so\[m+k=2\]so\[m=k=1\]but it gives\[mk=0\]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
you know k>m.. so from (k+m)(km)= 1*2 =>k+m =2 and km = 1 this doesnt have an integral solution..
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
why do i know k > m again?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
m^2 + 2 = k^2 both m and k are positive integers hence k>m
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
in layman language,, k^2  m^2 >0
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
when was it assumed that m and k are positive?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
what else do you mean by perfect square ?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i don't see how that relates.... the definition of a perfect square is x = k^2
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..
 one year ago

estudierBest ResponseYou've already chosen the best response.3
@mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
(2)^2 is also 4...
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
you take the absolute value
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
i mean you have to..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
see,, (k+m)(km) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take ve values,, negative * negative = positive ,,i.e. treated as absolute value..
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
...i don't think this is a valid proof....
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
another explanation might me x is a perfect square because its y^2 x = y^2 = (y)^2 sqrt(x) =  y  =  y 
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
since when did sqrt x become y
 one year ago

estudierBest ResponseYou've already chosen the best response.3
0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
so and i think that this difference of two will be the ,,key" of this proof of ...
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
@lgbasallote what is your opinion from this ,please ?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
my opinion is that this is some tricky algebra
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
how do you think it ???
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
because i can't think of a proof
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
how can you prove/disprove that the difference of two squares is 2...
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
their fields of specialty aren't exactly in proving
 one year ago

estudierBest ResponseYou've already chosen the best response.3
What's wrong with my proof?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
anyway... m^2 + 2 = k^2 k^2  m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2  (2y^2) = 2 4x^2  4y^2 = 2 then... 2x^2  2y^2 = 1 x^2  y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..
 one year ago

estudierBest ResponseYou've already chosen the best response.3
So, that doesn't mean it's invalid..
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
it does actually.....you can't do infinite substitutions to prove anything....
 one year ago

estudierBest ResponseYou've already chosen the best response.3
There is no infinity involved, all integer squares end in the numbers I gave.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
infinite loops involves infinity agree?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
No infinite loop either, just logic.
 one year ago

estudierBest ResponseYou've already chosen the best response.3
An integer ends in one of digits 0 to 9. The squares must end in etc....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
but we both know that setting values is illegal in proving
 one year ago

estudierBest ResponseYou've already chosen the best response.3
I am not setting any values.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
you're checking the squares of each number
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and though there is a pattern...there is still susbtitution involved
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and then you'll have to substitute forever to verify that the pattern really does not break
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
then infinity is involved
 one year ago

estudierBest ResponseYou've already chosen the best response.3
It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
so you do admit that you checked the 0 to 9 squares
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Anything else is impossible.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
...that's substitution
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
anyway...my point is...substitution isn't really allowed in proving
 one year ago

estudierBest ResponseYou've already chosen the best response.3
You will find that this fact is used a lot in many valid proofs in number theory.
 one year ago

estudierBest ResponseYou've already chosen the best response.3
OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
can you prove it? that it's a fact about squares?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Yes, all numbers end in 0 to 9 and so their squares end in .....QED.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
you didn't really demonstrate anything...you just repeated what you said
 one year ago

estudierBest ResponseYou've already chosen the best response.3
You asked for proof, I just gave u a proof.
 one year ago

estudierBest ResponseYou've already chosen the best response.3
If you deny my proof, provide a counterexample.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
since you try so hard to defend your proof, i'll accept it
 one year ago

estudierBest ResponseYou've already chosen the best response.3
That's very gracious, thank you:)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
do check on my algebra proof though please. i would like to know if that's the way contradiction works
 one year ago

estudierBest ResponseYou've already chosen the best response.3
OK, I will look at it now....
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
again i ask... how is k > m
 one year ago

estudierBest ResponseYou've already chosen the best response.3
(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[km=1\]\[k+m=2\]add them\[2k=3\] Contradiction
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
again i ask for the nth time @mukushla how is k > m
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
ok man i just want to answer ur question
 one year ago

estudierBest ResponseYou've already chosen the best response.3
"2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
well because 8  4 is not 2 but it's even minus even
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
OK, so finally, the algebra Given n=k^2 > (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
i thought they are positive :) sorry
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
why (k+1)^2 @estudier ?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
I already, said, that's the next square....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
is this still related to my proof?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
? n is k^2, what is the next square?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
no problem ... this is better\[k^2m^2=2\]\[(km)(k+m)=2\]so\[km=1\]\[k+m=2\]because this time im sure that \(k>m\) ...now add them\[2k=3\]Contradiction and we are done
 one year ago

estudierBest ResponseYou've already chosen the best response.3
No, this proof Given n=k^2 > (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
are you allowed to do that? @mukushla
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
why not...for any real number \[a^2=a^2\]
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Given n=k^2 > (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED
 one year ago

estudierBest ResponseYou've already chosen the best response.3
That's 2 proofs I have given now...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Think about it....:)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
in the meanwhile... @mukushla why k  m = 1?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
because\[km<k+m\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
how does that make k  m = 1
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
u have 2 positive integers with ab=2 and a<b what are a and b....clearly 1 and 2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
who said those are positive?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
something feels wrong...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
anyway...where did you get 2k > 3
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
ok...think on what i said...because i believe thats the shortest way to prove ur statement
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
adding\[km=1\]and\[k+m=2\]
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Going back to your question about 2, u can just adjust it to difference of 2q...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
im really skeptical about this... especially that k^2  m^2 = 2
 one year ago

estudierBest ResponseYou've already chosen the best response.3
You were worried about the validity of your statement, change it to 2q instead of just 2.
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Or 2p, or whatever letter u like best.
 one year ago

estudierBest ResponseYou've already chosen the best response.3
1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
oh. when i said difference of two numbers equal to 2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and @mukushla how did you get k > m
 one year ago

estudierBest ResponseYou've already chosen the best response.3
No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Yes, the algebra is completely general...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...
 one year ago

estudierBest ResponseYou've already chosen the best response.3
...which is greater than n+3, right?
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Number theorists don't pay attention to 0
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....
 one year ago

estudierBest ResponseYou've already chosen the best response.3
"because i know nothing about it" But you do now:)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
it's not like it's limited to the 99s....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.
 one year ago

estudierBest ResponseYou've already chosen the best response.3
Yes, that's similar to the one I gave...
 one year ago

estudierBest ResponseYou've already chosen the best response.3
"number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number
 one year ago

estudierBest ResponseYou've already chosen the best response.3
http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?
 one year ago

jhonyy9Best ResponseYou've already chosen the best response.0
so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 2 so m1,2 = +/ sqrt(x^2 2)
 one year ago
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